Add nonconvex qp example (#335)

* Add nonconvex qp example

* Fix

* Expand explanation

Author: blegat

docs/src/tutorials/Polynomial Optimization/qp.jl

@@ -0,0 +1,75 @@
+# # Nonconvex QP
+
+#md # [![](https://mybinder.org/badge_logo.svg)](@__BINDER_ROOT_URL__/generated/Polynomial Optimization/qp.ipynb)
+#md # [![](https://img.shields.io/badge/show-nbviewer-579ACA.svg)](@__NBVIEWER_ROOT_URL__/generated/Polynomial Optimization/qp.ipynb)
+# **Adapted from**: Section 2.2 of [F99], Table 5.1 of [Las09]
+#
+# [F99] Floudas, Christodoulos A. et al.
+# *Handbook of Test Problems in Local and Global Optimization.*
+# Nonconvex Optimization and Its Applications (NOIA, volume 33).
+#
+# [Las09] Lasserre, J. B.
+# *Moments, positive polynomials and their applications*
+# World Scientific, **2009**.
+
+# ## Introduction
+
+# Consider the nonconvex Quadratic Program (QP) [F99, Section 2.2]
+# that minimizes the *concave* function $c^\top x - x^\top Qx / 2$
+# over the polyhedron obtained by intersecting the hypercube $[0, 1]^5$
+# with the halfspace $10x_1 + 12x_2 + 11x_3 + 7x_4 + 4x_5 \le 40$.
+
+using Test #src
+
+using LinearAlgebra
+c = [42, 44, 45, 47, 47.5]
+Q = 100I
+
+using DynamicPolynomials
+@polyvar x[1:5]
+p = c'x - x' * Q * x / 2
+using SumOfSquares
+K = @set x[1] >= 0 && x[1] <= 1 &&
+         x[2] >= 0 && x[2] <= 1 &&
+         x[3] >= 0 && x[3] <= 1 &&
+         x[4] >= 0 && x[4] <= 1 &&
+         x[5] >= 0 && x[5] <= 1 &&
+         10x[1] + 12x[2] + 11x[3] + 7x[4] + 4x[5] <= 40
+
+# We will now see how to find the optimal solution using Sum of Squares Programming.
+# We first need to pick an SDP solver, see [here](https://jump.dev/JuMP.jl/v1.12/installation/#Supported-solvers) for a list of the available choices.
+
+import Clarabel
+solver = Clarabel.Optimizer
+
+# A Sum-of-Squares certificate that $p \ge \alpha$ over the domain `S`, ensures that $\alpha$ is a lower bound to the polynomial optimization problem.
+# The following function searches for the largest lower bound and finds zero using the `d`th level of the hierarchy`.
+
+function solve(d)
+    model = SOSModel(solver)
+    @variable(model, α)
+    @objective(model, Max, α)
+    @constraint(model, c, p >= α, domain = K, maxdegree = d)
+    optimize!(model)
+    println(solution_summary(model))
+    return model
+end
+
+# The first level of the hierarchy does not give any lower bound:
+
+model2 = solve(2)
+nothing # hide
+@test termination_status(model2) == MOI.INFEASIBLE #src
+
+# Indeed, as the constraints have degree 1 and their multipliers are SOS
+# so they have an even degree, with `maxdegree` 2 we can only use degree 0
+# multipliers hence constants. The terms of maximal degree in resulting
+# sum will therefore only be in `-x' * Q * x/2` hence it is not SOS whatever
+# is the value of the multipliers. Let's try with `maxdegree` 3 so that the
+# multipliers can be quadratic.
+# This second level is now feasible and gives a lower bound of `-22`.
+
+model3 = solve(3)
+nothing # hide
+@test objective_value(model3) ≈ -22 rtol=1e-4 #src
+@test termination_status(model3) == MOI.OPTIMAL #src